[tex]~~~~~~\tan^{-1}(xy) = \sin^{-1}(4x+4y)\\\\\\\implies \dfrac d{dx} \left[ \tan^{-1}(xy)\right] = \dfrac d{dx} \sin^{-1}\left( 4x+4y\right)\\ \\\\\implies \dfrac{1}{1+(xy)^2} \left[ \dfrac{d}{dx} (xy)\right] = \dfrac{1}{\sqrt{1-(4x+4y)^2}} \cdot \dfrac{d}{dx}(4x+4y)\\\\\\\implies \dfrac{1}{1+x^2 y^2} \left(x\dfrac{dy}{dx}+y \right)=\dfrac{1}{\sqrt{1-(4x+4y)^2}} \left(4+ 4 \dfrac{dy}{dx} \right)\\\\\\\\[/tex]
[tex]\implies \left(\dfrac{x}{1+x^2y^2} \right) \dfrac{dy}{dx} + \dfrac{y}{1+x^2y^2}=\dfrac{1}{\sqrt{1-(4x+4y)^2}} \left(4+ 4 \dfrac{dy}{dx} \right)\\[/tex]
[tex]\\\\\implies 0 \cdot \dfrac{dy}{dx} +0 = \dfrac{1}{\sqrt{1 -0}} \left(4+4 \dfrac{dy}{dx} \right) ~~~~~~~~~;\left[ \text{at}~~ (0,0) \right]\\\\\\\implies 4+4 \dfrac{dy}{dx} = 0\\\\\\\implies \dfrac{dy}{dx} = -\dfrac 44\\\\\\\implies \dfrac{dy}{dx} = -1\\\\\text{So, the slope of the tangent line at (0,0) is}~ -1\\ \\\\\text{The equation of tangent line,}\\\\~~~~~~y-0 = -1(x-0)\\\\\implies y=-x[/tex]
